# Are there other, non-probabilistic ways to calculate: $\lim_{n\to\infty}\frac{1}{n}\ln\sum_{m>n\alpha}\frac{(n\lambda)^m}{m!}$?

In section five of this nice exposition of moment generating functions, we prove the following theorem:

Take an i.i.d sequence of random variables $$(X_n:\Omega\to\Bbb R)_{n\in\Bbb N}$$ whose common cumulant generating function $$\Lambda$$ is finite in a neighbourhood of the origin (the value $$+\infty$$ is permitted in general). Let $$X:=X_1$$.

Define $$\Lambda^\ast$$ to be the "Fenchel-Legendre transform" of $$\Lambda$$: $$\Lambda^\ast:\Bbb R\to\overline{\Bbb R},\,x\mapsto\sup_{t\in\Bbb R}(x\cdot t-\Lambda(t))$$And define for every $$n\in\Bbb N$$ the empirical means: $$Y_n:=\frac{1}{n}\sum_{j=1}^nX_j:\Omega\to\Bbb R$$

If $$\alpha>\mathbb{E}(X)$$ and $$\mathrm{Pr}(X>\alpha)>0$$ then $$0<\Lambda^\ast(\alpha)<\infty$$ and we get the asymptotics of the "large deviations": $$\frac{1}{n}\ln\mathrm{Pr}(Y_n>\alpha)\overset{n\to\infty}{\longrightarrow}-\Lambda^\ast(\alpha)=\inf_{t>0}(\Lambda(t)-\alpha\cdot t)$$With convergence from below.

The proof is - to me - a fairly complex and delicate application of probability theory. I don't yet know enough to follow it all the way through. I wanted to "see it in action" so I explicitly calculated $$\Lambda^\ast$$ for the Poisson distribution.

Let's fix some $$\lambda>0$$. If we take $$(X_n)_n$$ which follow the Poisson distribution with parameter $$\lambda$$, it is easy to check that $$\Lambda(t)=\lambda(e^t-1)$$ for all $$t$$. Some basic calculus optimisation will find: $$\Lambda^\ast(\alpha)=\lambda+\alpha(\ln(\alpha\cdot\lambda^{-1})-1)$$Whenever $$\alpha>\lambda$$.

So the theorem predicts that: $$\lim_{n\to\infty}\frac{1}{n}\ln\left(\sum_{m>n\alpha}\frac{(n\lambda)^m}{m!}e^{-n\lambda}\right)=-\lambda+\alpha(1-\ln(\alpha\cdot\lambda^{-1}))$$A little rearranging brings that to: $$\tag{\ast}\forall\,\,0<\lambda<\alpha:\quad\quad\quad\lim_{n\to\infty}\frac{1}{n}\ln\left(\sum_{m>n\alpha}\frac{(n\lambda)^m}{m!}\right)=\alpha(1-\ln(\alpha\cdot\lambda^{-1}))$$

Which seems highly nontrivial to do by other means. I am very curious to see whether or not anyone on the site has the expertise to supply a "real analytic" proof of this identity. I know that the probabilistic proof is just a real analytic proof with a particular flavour, but I mean to ask if there are other ways of doing this which aren't purely motivated by ideas from probability or measure theory (the proof seems to involve a "change-of-measure" trick, which is a new one on me). Or, if this special Poisson case admits a simpler probabilistic proof, that would be interesting to see too.

I think this special case is an interesting problem. Since I don't fully understand the proof I've already seen, I reckon that verifying $$(\ast)$$ is way above my paygrade. I hope to learn from your answers :)

Just one thing I can helpfully observe:

By the Stolz-Cesaro theorem, it would suffice to show that (although I do not know if this is true): $$\lim_{n\to\infty}\frac{\sum_{m>(n+1)\alpha}\frac{((n+1)\lambda)^m}{m!}}{\sum_{m>n\alpha}\frac{(n\lambda)^m}{m!}}=\left(\frac{\lambda e}{\alpha}\right)^\alpha$$

Oh, and a related cool (I think) expression can be deduced by applying the same technique to a sequence of binomial variable variables with parameters $$n,1/2$$:

If $$n\in\Bbb N$$ and $$\frac{n}{2}<\alpha then: $$\lim_{m\to\infty}\frac{1}{m}\ln\left(\sum_{k>m\alpha}\binom{mn}{k}\right)=n\ln n-\alpha\ln\alpha-(n-\alpha)\ln(n-\alpha)$$

More generally, if $$0, $$n\in\Bbb N$$ and $$np<\alpha: $$\lim_{m\to\infty}\frac{1}{m}\ln\left(\sum_{k>m\alpha}\binom{mn}{k}\left(\frac{p}{1-p}\right)^k\right)=\alpha\ln\left(\frac{p}{1-p}\right)+n\ln n-\alpha\ln\alpha-(n-\alpha)\ln(n-\alpha)$$

This also seems quite nontrivial without this “master theorem”. I would be equally interested to see a proof of that by other means.

• Did you try writing the integral formula for the error term in the Taylor expansion of $e^x$ of degree $m \approx n\alpha$ and setting $x=n \lambda$? Perhaps that bound might suffice. Mar 18 at 15:47
• @MathWonk No I haven’t tried this. I’ll think on it. Thanks for any suggestions Mar 19 at 23:52

Assume $$0<\lambda<\alpha$$. By $$(8.4.9)$$ and $$(8.6.3)$$, we have $$\tag{1} \hspace{-20pt}\sum\limits_{m > n\alpha } {\frac{{(n\lambda )^m }}{{m!}}} = {\rm e}^{n\lambda } P(n\alpha + 1,n\lambda ) = \frac{{(n\lambda )^{n\alpha + 1} {\rm e}^{n\lambda } }}{{\Gamma (n\alpha + 1)}}\int_0^{ + \infty } {\exp ( - n(\alpha t + \lambda {\rm e}^{ - t} )){\rm e}^{ - t} {\rm d}t} .$$ The saddle point $$t = \log (\lambda /\alpha ) < 0$$ is outside the range of integration and the minimum of $$t\mapsto \alpha t + \lambda {\rm e}^{ - t}$$ occurs at $$t=0$$. Whence, by Laplace's method, $$\int_0^{ + \infty } {\exp ( - n(\alpha t + \lambda {\rm e}^{ - t} )){\rm e}^{ - t} {\rm d}t} \sim \frac{1}{{\alpha - \lambda }}\frac{{{\rm e}^{ - n\lambda } }}{n}$$ as $$n\to+\infty$$. Hence by $$(1)$$ and Stirling's formula, $$\sum\limits_{m > n\alpha } {\frac{{(n\lambda )^m }}{{m!}}} \sim \frac{\lambda }{{\alpha - \lambda }}\sqrt {\frac{{2\pi n}}{\alpha }} \left( {\frac{{\lambda {\rm e}}}{\alpha }} \right)^{n\alpha }$$ as $$n\to+\infty$$. Finally, $$\frac{1}{n}\log \!\bigg( {\sum\limits_{m > n\alpha } {\frac{{(n\lambda )^m }}{{m!}}} } \bigg) \sim \alpha \log\! \bigg( {\frac{{\lambda {\rm e}}}{\alpha }} \bigg) = \alpha (1 - \log (\alpha /\lambda ))$$ as $$n\to+\infty$$.
• Very clean. One thing: I don’t think I saw it there - what are $P$ and $Q$? Mar 20 at 8:38
• If you click on the link it will define all the notation appearing in the formula. The $P(a,z)$ is the normalised incomplete gamma function: $\gamma(a,z)/\Gamma(a)$. DLMF is a useful website with lots of information.
• Click on the $i$ icon on the right-hand side of the section title $\S2.3(\text{iii})$ and it will tell you with a link to the reference: "See Olver (1997b, pp. 80–88)."